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1
M602: Methods and Applications of Partial Differential Equations. Final TEST, May, 2014. Notes,
books, and calculators are not authorized. Show all your work in the blank space you are given on
the exam sheet. Always justify your answer. Answers with no justification will not be graded.
Here are some formulae that you may want to use:
Z +∞
Z +∞
def 1
f (x)eiωx dx,
F −1 (f )(x) =
f (ω)e−iωx dω,
F(f )(ω) =
2π −∞
−∞
(
1 if |x| ≤ λ
1 sin(λω)
F(Sλ (x)) =
, where Sλ (x) =
π
ω
0 otherwise
F(f ∗ g) = 2πF(f )F(g),
(1)
(2)
The implicit representation of the solution to the equation ∂t v + ∂x q(v) = 0, v(x, 0) = v0 (x), is
X(s, t) = q 0 (v0 (s))t + s;
v(X(s, t), t) = v0 (s).
(3)
Question 1: Let y(x, t) = x cos(2t + log(|x|)). Compute ∂tt y + x2 ∂xx y − x∂x y + 6y.
Solution: This exercise is meant to check whether you understand the notion of partial derivatives and the chain rule
∂tt y(x, t) = −4x cos(2t + log(|x|)) = −4y(x, t),
∂x y(x, t) = cos(2t + log(|x|)) − x sin(2t + log(|x|))
1
= cos(2t + log(|x|)) − sin(2t + log(|x|)),
x
1
1
∂xx y(x, t) = − sin(2t + log(|x|)) − cos(2t + log(|x|))
x
x
In conclusion
∂tt y + x2 ∂xx y − x∂x y + 6y = −4x cos(2t + log(|x|)) − x sin(2t + log(|x|))
− x cos(2t + log(|x|)) − x cos(2t + log(|x|)) + x sin(2t + log(|x|)) + 6x cos(2t + log(|x|))
= 0,
that is to say, y(x, t) solve the PDE ∂tt y + x2 ∂y − x∂x y + 6y = 0.
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2
Question 2: Let µ and c be two positive numbers. Consider the following system of coupled partial differential
equations with dependent variables ρ(x, t) and u(x, t): ∂t ρ + ∂x u = 0, ∂t u − µ∂xx u + c2 ∂x ρ = 0, x ∈ (0, L), t > 0,
with initial data ρ(x, 0) = ρ0 , u(x, 0) = u0 , and boundary conditions u(0, t) = u(L, t) = 0. (i) Show that the
RL
quantity E(t) = 0 12 (ρ2 (x, t) + c−2 u(x, t))dx decreases in time. (Hint: Use the energy argument: Use ρ for the
first equation, u/c2 for the second, add the results, do not panic and proceed as usual.)
Solution: We proceed as in the hint. We multiply the first equation by ρ and integrate over the domain,
Z
0=
L
Z
L
(ρ(x, t)∂t ρ(x, t) + ρ(x, t)∂x u(x, t))dx =
0
d
=
dt
0
L
Z
0
1 2
ρ (x, t)dx +
2
Z
1
(∂t ( ρ2 (x, t)) + ρ(x, t)∂x u(x, t))dx
2
L
ρ(x, t)∂x u(x, t)dx.
0
We proceed similarly for the second equation. We multiply the second equation by c−2 u and integrate over the domain,
L
Z
(c−2 u(x, t)∂t u(x, t) − µc−2 u(x, t)∂xx u(x, t) + u(x, t)∂x ρ(x, t))dx
0=
0
We integrate by parts the viscous term and we obtain −
to the boundary conditions u(0, t) = u(L, t) = 0,
Z
RL
0
µc−2 u(x, t)∂xx u(x, t)dx =
RL
0
µc−2 (∂x u(x, t))2 dx, owing
L
1
(c−2 ∂t ( u2 (x, t)) + µc−2 (∂x u(x, t))2 + u(x, t)∂x ρ(x, t))dx
2
0
Z
Z L
d L 1 −2 2
=
c u (x, t)dx +
(µc−2 (∂x u(x, t))2 + u(x, t)∂x ρ(x, t))dx.
dt 0 2
0
0=
Adding the two equations together and using the product rule gives
0=
=
=
d
dt
Z
d
dt
Z
d
dt
Z
L
0
L
0
0
L
Z L
1 −2 2
c u (x, t)dx +
(µc−2 (∂x u(x, t))2 + u(x, t)∂x ρ(x, t))dx
0
0 2
0
Z L
Z L
1 2
(ρ (x, t) + c−2 u2 (x, t))dx +
µc−2 (∂x u(x, t))2 dx +
(ρ(x, t)∂x u(x, t) + u(x, t)∂x ρ(x, t))dx
2
0
0
Z L
Z L
1 2
(ρ (x, t) + c−2 u2 (x, t))dx +
µc−2 (∂x u(x, t))2 dx +
∂x (ρ(x, t)u(x, t))dx.
2
0
0
1 2
ρ (x, t)dx +
2
Z
L
ρ(x, t)∂x u(x, t)dx +
d
dt
Z
L
RL
The Fundamental Theorem of Calculus gives 0 ∂x (ρ(x, t)u(x, t))dx = 0, owing to the boundary conditions on u. In
conclusion
Z
Z L
d L1 2
(ρ (x, t) + c−2 u2 (x, t))dx = −
µc−2 (∂x u(x, t))2 dx ≤ 0,
dt 0 2
0
RL 1 2
−2
which proves that the quantity E(t) = 0 2 (ρ (x, t) + c u(x, t))dx decreases in time.
(ii) What how does E(t) behaves when µ = 0?
Solution: The above computation shows that
µ = 0.
d
dt E(t)
= 0 when µ = 0; as a result, E(t) is constant in time when
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3
Solve the PDE ∂t u + 4∂x u + 12u = 0 in Ω = {(x, t) ∈ R2 | x ≥ 0, x + 11t ≥ 0} with boundary conditions
u(x, 0) = x + 4, u(−11t, t) = t + 4 for t > 0.
Solution: (1) The boundary of Ω, say Γ, is parametrized as follows: Γ = {(xΓ (s), tΓ (s)) | s ∈ R} where
(
(
−s if s < 0,
11s if s < 0,
tΓ (s) =
xΓ (s) =
0
if s > 0.
s
if s > 0.
(2) The characteridtics are defined by the ODE
∂t X(s, t) = 4,
X(s, tΓ (s)) = xΓ (s).
The solution is X(s, t) = 4(t − tΓ (s)) + xΓ (s).
(3) We make a change of variable: φ(s, t) = u(X(s, t), t) and compute ∂t φ(s, t),
∂t φ(s, t) = ∂t u(X(s, t), t) + ∂x u(X(s, t), t)∂t X(s, t) = ∂t u(X(s, t), t) + 4∂x u(X(s, t), t)
= −12u(X(s, t), t) = −12φ(s, t),
with initial data φ(s, tΓ(s) ) = u(xΓ(s) , tΓ(s) ). The solution is
φ(s, t) = φ(s, tΓ(s) )e−12(t−tΓ(s) ) = u(xΓ(s) , tΓ(s) )e−12(t−tΓ(s) ) .
i.e., the implicit representation of the solution is
u(X(s, t), t) = u(xΓ(s) , t)e−12(t−tΓ(s) ) ,
X(s, t) = 4(t − tΓ (s)) + xΓ (s).
(4) The explicit representation is obtained as follows:
Case 1 (s < 0): X(s, t) = 4(t + s) + 11s, then s = (X − 4t)/15 and
u(x, t) = u(11s, −s)e−12(t+s) = (−s + 4)e−12(t+s) = (4 −
= (4 +
(4t − x) −4 11t+x
5
)e
,
15
(x − 4t) −12(t+ x−4t )
15
)e
15
if x < 4t.
Case 2 (s > 0): X(s, t) = 4t + s, then s = X − 4t, and
u(x, t) = u(s, 0)e−12t = (s + 4)e−12t = (4 + x − 4t)e−12t
if 4t < x.
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Question 3: Consider the telegraph equation ∂tt u + 2α∂t u + α2 u − c2 ∂xx u = 0 with α ≥ 0, u(x, 0) = 0, ∂t u(x, 0) =
g(x), x ∈ R, t > 0 and boundary condition at infinity u(±∞, t) = 0. Solve the equation by the Fourier transform
technique. (Hint: the solution to the ODE φ00 (t) + 2αφ0 (t) + (α2 + λ2 )φ(t) = 0 is φ(t) = e−αt (a cos(λt) + b sin(λt))
Solution: Applying the Fourier transform with respect to x to the equation, we infer that
0 = ∂tt F(u)(ω, t) + 2α∂t F(u)(ω, t) + α2 F(u)(ω, t) − c2 (−iω)2 F(u)(ω, t)
= ∂tt F(u)(ω, t) + 2α∂t F(u)(ω, t) + (α2 + c2 ω 2 )F(u)(ω, t)
Using the hint, we deduce that
F(u)(ω, t) = e−αt (a(ω) cos(ωct) + b(ω) sin(ωct)).
The initial condition implies that a(ω) = 0 and F(g)(ω) = ωcb(ω); as a result, b(ω) = F(g)(ω)/(ωc) and
sin(ωct)
.
ωc
F(u)(ω, t) = e−αt F(g)
Then using (2), we have
F(u)(ω, t) =
π −αt
e F(g)F(Sct ).
c
The convolution theorem implies that
u(x, t) = e
−αt
1
1
g ∗ Sct = e−αt
2c
2c
Z
∞
g(y)Sct (x − y)dy.
−∞
Finally the definition of Sct implies that Sct (x − y) is equal to 1 if −ct < x − y < ct and is equal zero otherwise, which
finally means that
Z x+ct
1
u(x, t) = e−αt
g(y)dy.
2c x−ct
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5
Question 4: Consider the equation −∂x ((1 + x)∂x u(x)) + ∂x u(x) = f (x), x ∈ (0, 1) with u(0) = α and −2∂x u(1) +
u(1) = β. Let G(x, x0 ) be the Green’s function. (i) Give the integral representation of u(x0 ) for all x0 ∈ (0, 1) in
terms of G, f , α and β and give the equation and boundary conditions that G must satisfy. Do not compute G at
this question. (Hint: The differential operator is not self-adjoint. You should find G(0, x0 ) = 0, ∂x G(1, x0 ) = 0).
Solution: We multiply the PDE by G(x, x0 ) and integrate by parts,
1
Z
1
Z
((1 + x)∂x u(x)∂x G(x, x0 ) − u(x)∂x G(x, x0 )) dx + [(−(1 + x)∂x u(x) + u(x))G(x, x0 )]10
f (x)G(x, x0 )dx =
0
0
1
Z
((1 + x)∂x u(x)∂x G(x, x0 ) − u(x)∂x G(x, x0 )) dx + βG(1, x0 ) − (−∂x u(0) + α)G(0, x0 ).
=
0
Since ∂x u(0) is not known, we must have G(0, x0 ) = 0. Then
Z
1
Z
1
(−u(x)∂((1 + x)∂x G(x, x0 )) − u(x)∂x G(x, x0 )) dx + βG(1, x0 ) + [(1 + x)u(x)∂x G(x, x0 )]10
f (x)G(x, x0 )dx =
0
0
Z
1
−u(x) (∂((1 + x)∂x G(x, x0 )) + ∂x G(x, x0 )) dx + βG(1, x0 ) + 2u(1)∂x G(1, x0 ) − α∂x G(0, x0 ).
=
0
Since u(1) is not known, we must have ∂x G(1, x0 ) = 0. Finally G must satisfy
−∂x ((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = δ(x − x0 ),
G(0, x0 ) = 0,
∂x G(1, x0 ) = 0,
and we have the following representation for u(x0 ),
Z
1
f (x)G(x, x0 )dx − βG(1, x0 ) + α∂x G(0, x0 ).
u(x0 ) =
0
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(ii) Compute G(x, x0 ) such that −∂x ((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = δ(x − x0 ), G(0, x0 ) = 0, ∂x G(1, x0 ) = 0,
for all x, x0 ∈ (0, 1). (Hint: observe that (1 + x)φ0 (x) + φ(x) = ((1 + x)φ(x))0 .)
Solution: For all x 6= x0 we have
−∂((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = −∂((1 + x)∂x G(x, x0 ) + G(x, x0 )) = 0
Using the hint, this implies that
(1 + x)∂x G(x, x0 ) + G(x, x0 ) = ∂x ((1 + x)G(x, x0 )) = a,
In conclusion
(
G(x, x0 ) =
ax+b
1+x
cx+d
1+x
if x ≤ x0
if x0 ≤ x.
The boundary condition at 0 gives
G(0, x0 ) = 0 = b,
ax
1+x
i.e., b = 0, G(x, x0 ) =
if x ≤ x0 . The boundary condition at 1 gives
∂x G(0, x0 ) =
c(1 + 0) − (c × 0 + d) × 1
= 0,
(1 + 0)2
i.e., c = d. G(x, x0 ) = c, if x0 ≤ x. We need to impose the continuity of G(x, x0 ) at x0 ,
ax0
= c,
1 + x0
which gives ax0 = c(1 + x0 ). The jump condition gives
Z
x0 +
1 = lim
→0
x0 −
= (1 + x0 )(
+
(−∂x ((1 + x)∂x G(x, x0 )) + ∂x G(x, x0 )) dx = (1 + x0 )(∂x G(x−
0 , x0 ) − ∂x G(x0 , x0 ))
a(1 + x0 ) − ax0
a
)=
.
2
(1 + x0 )
1 + x0
In conclusion a = (1 + x0 ) and c = x0 , Then
(
G(x, x0 ) =
x(1+x0 )
1+x
x0
if x ≤ x0
if x0 ≤ x.
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Question 5: Consider the conservation equation ∂t ρ + ∂x (sin( π2 ρ)) = 0, x ∈ R, t > 0, with initial data ρ0 (x) = 1
if x < 0 and ρ0 (x) = 0 if x > 0. Draw the characteristics and give the explicit representation of the solution.
Solution: The implicit representation of the solution to the equation ∂t ρ + ∂x q(ρ) = 0, ρ(x, 0) = ρ0 (x), is
X(s, t) = q 0 (ρ0 (s))t + s;
ρ(X(s, t), t) = ρ0 (s).
(4)
The explicit representation is obtained by expressing s in terms of X and t.
Case 1: s < 0, we have ρ0 (s) = 1, q 0 (ρ0 (s)) = π2 cos( π2 ) = 0, which implies X = s. Then
ρ(x, t) = 1 if x < 0.
Case 2: 0 < s, we have ρ0 (s) = 0, q 0 (ρ0 (s)) =
π
2
cos(0) =
π
2,
ρ(x, t) = 0 if
X=
π
2t
+ s, which means s = X − π2 t. Then
π
t < x.
2
Case 3: s = 0. Note that there is no characteristics in the region 0 < x <
wave in this region.
π
2 t;
this means that there is an expansion
The solution is given by setting s = 0 in the expression defining te characteristics: X = q 0 (ρ0 )t + 0 =
the constraint that 0 < ρ0 (s) < 1. This means that
ρ0 =
2
2X
arccos(
),
π
πt
for
0≤X≤
π
2
cos( π2 ρ0 )t with
π
t.
2
Finally
ρ(x, t) =


1
2
π

0
arccos( 2x
πt ),
if x < 0,
0 ≤ x ≤ π2 t,
if π2 t < x.
(5)
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Question 6: Consider the conservation equation with flux q(ρ) = ρ3 . Assume that the initial data is ρ0 (x) = 2, if
x < 0, ρ0 (x) = 1, if 0 < x < 1, and ρ0 (x) = 0, if 1 < x. (i) Draw the characteristics
Solution: There are three families of characteristics.
Case 1: s < 0, X(s, t) = 12t + s. In the x-t plane, these are lines with slope
1
12 .
Case 2: 0 < s < 1, X(s, t) = 3t + s. In the x-t plane, these are lines with slope 13 .
Case 3: 1 < s, X(s, t) = s. In the x-t plane, these are vertical lines.
One shock forms between the two black characteristics and another forms between the two red characteristic (see figure).
(ii) Describe qualitatively the nature of the solution.
Solution: We have two shocks moving to the right. One shock forms between the two black characteristics and another
forms between the two red characteristic (see figure).
(iii) When and where does the left shock catch up with the right one?
Solution: The speeds of the shocks are
dx1 (t)
23 − 1
=
= 7,
dt
2−1
and
dx2 (t)
1−0
=
= 1.
dt
1−0
The location of the left shock at time t is x1 (t) = 7t and that of the right shock is x2 (t) = t + 1. The two shocks are
at the same location when 7t = t + 1, i.e., t = 16 ; the two shocks merge at x = 67 .
(iv) What is the speed of the shock once the two shocks have merged and what is the position of the shock as a
function of time?
Solution: When the shocks have merged the left state is ρ = 2 and the right state is ρ = 0; as a result the speed of
the shock is
23 − 0
dx3 (t)
=
= 4,
dt
2−0
and the shock trajectory is x3 (t) = 4(t − 16 ) + 76 .
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(v) Draw precisely all the characteristics of the solution.
9
Solution: The three shocks are shown in color.
Question 7: Consider the following Cauchy-Euler problem: Let f be a smooth function in [0, π2 ] and find u such
π
π
that 13u − 5x∂x u + x2 ∂xx u = f (x), x ∈ (1, e 2 ), u(1) = 0, u(e 2 ) = 0. (i) Does this problem have a unique solution.
(Hint: The solution to the homogeneous equation (i.e., f = 0) is v(x) = ax3 cos(2 log(|x|)) + bx3 sin(2 log(|x|)).)
Solution: Let us compute the null space of the operator
π
L : {v ∈ C 2 (0, π)| v( ) = 0, v(0) = 0} 3 u 7−→ 13u − 5x∂x u + x2 ∂xx u ∈ C 0 (0, π).
2
Let N (L) be the null space. Let v ∈ N (L), then
13v − 5x∂x v + x2 ∂xx v = 0
which means that v = ax3 cos(2 log(|x|)) + bx3 sin(2 log(|x|)). The boundary conditions imply that
0 = a cos(2 log(1)) + bx3 sin(2 log(1)) = a,
π
0 = b(π/2)3 sin(2 log(e 2 )) = b(π/2)3 sin(π),
a = 0 and b is an arbitrary real number; as a result N (L) = span(x3 sin(2 log(|x|))), i.e., N (L) is the one-dimensional
vector space spanned by the function x3 sin(2 log(|x|)). Assuming that the problem has a solution, it cannot be unique.
We are in the second case of Fredholm’s alternative. Once N (LT ) is charaterized, the above problem has a solution
Rπ
only if 12 vf (x)dx = 0 for all v ∈ N (LT ). Question: What is N (LT )?